Question

The centre of the circle $x=2+3\cos \theta ,y=3\sin \theta -1$ is

• A. $(3,3)$
• B. $(2,-1)$
• C. $(-2,1)$
• D. $(-1,2)$

Answer 1

Answer: (B)

$(2,-1)$

The given parametric equations $x=2+3\cos \theta$ and $y=3\sin \theta -1$ can be rewritten as:

$x=2+3\cos \theta \Rightarrow 3\cos \theta =x-2\Rightarrow \cos \theta =\frac{x-2}{3}$ ..............(1)

$\Rightarrow y=3\sin \theta -1\Rightarrow 3\sin \theta =y+1\Rightarrow \sin \theta =\frac{y+1}{3}$..........(2)

Now since ${\sin }^2\theta +{\cos }^2\theta =1$, substitute the values from equations (1) and (2):

${\sin }^2\theta +{\cos }^2\theta =1\Rightarrow {\left(\frac{x-2}{3}\right)}^2+{\left(\frac{y+1}{3}\right)}^2=1\Rightarrow {(x-2)}^2+{(y+1)}^2={\left(3\right)}^2$

The general equation of circle with centre $(h,k)$ and radius $r$ is given by ${(x-h)}^2+{(y-k)}^2={\left(r\right)}^2$.

Thus, from the equation ${(x-2)}^2+{(y+1)}^2={\left(3\right)}^2$, we get that $h=2$ and $k=-1$.

Hence, the centre of the circle is $(2,-1)$.